For all $M$ on the boundary of $A$, let $A_M$ be the homothetic of $A$ with respect to the homothty $h_M$ of pole $M$ and ratio $\frac 2 3 $. Using the convexity of $A$, we deduce that $A_M$ is convex and contained in $A$ for all such point $M$. Using the convexity of $A$ and $A_M$, it is easy to verify that is $M,N,P$ are three points on the boundary of $A$ and $G$ is the center of gravity of $MNP$ then $G$ belongs to each of the sets $A_M,A_N,A_P$. It follows that the family of the sets $A_M$ is a set of bounded convex sets such that any three have a common point. From Helly's theorem, we deduce that there exists a point $O$ which belongs to all the sets $A_M$ where $M$ is a point of the boundary of $A$. Now, let's consider any line $d$ which passes through $O$. This line meet the boundary of $A$ in two points $X,X'$. Then, from above $O$ belongs to the segment $XX''$ where $X'' = h_X(X')$, and to the segment $XX'$. Thus, $OX \leq XX'' = \frac 2 3 XX'$ so that $OX' \geq \frac 1 3 XX'$. Therefore $\frac {OX } {OX'} \leq 2$. The other inequality is obtained by interchanging $X$ and $X'$. Pierre.

It's very similar to this problem. In fact, in order to prove that, I proved a very similar lemma (although weaker).

what is HELLYS THEOREM?

Given a collection of convex sets in $\mathbb R^{n}$ s.t. each $n+1$ have non-void intersection, all of them have non-void intersection.

Another solution that is very beautiful. If A and B are two convex figures. Then there is atranslation that takes A in B if and only if Every 3 points of A could be taken in B with a translation. Proof of this is easily with Helly theorem.

A is a convex set in plane so it is compact. So, we can set a triangle which the size is maximum. If we set $ O $ the centroid of it, we are done.

Here is another thread dedicated to this problem. One can see that the center of gravity(the centroid) of $A$ satisfies the requirement.

a nice problem here's my solution: we will use $S$ to refer to the convex set take a triangle $ABC$ with vertices in the boundary of $S$ with maximum area note that if we construct the parallel from $ A$ to $BC$ let it $\ell_a$ similarly $\ell_c$ and $\ell_b$ then the triangle from $\ell_a$ $\ell_c$ $\ell_b$ contains $S$ (let it $T$) take $O$ the centroid of $ABC$ take a line throw $O$ cuts $S$ in $X,Y$, $ABC$ at $X',Y'$ and $T$ at $X'',Y''$ (with $X',X''$ closer to $X$) $2 \ge \frac{OX''}{OY'} \ge \frac{OX}{OY} \ge \frac{OX'}{OY''} \ge \frac{1}{2}$ and we win

It can be proved immediately from Minkovski's theorem. It states that for convex sets in $R^d$ we can find a point $O$ such that for every line $XOX'$ we have $\dfrac{XO}{X'O} \leq d$